1. A body having moment of inertia of 30 kg m2 is rotating at 210 RPM and mashes with another body at rest having M.I. of 40 kg m2. The resultant speed after meshing will be
a) 90 RPM
b) 100 RPM
c) 80 RPM
d) none of the mentioned
Explanation: Since moment is conserved, therefore,
30 x 210 = 40 x Resultant speed
Resultant speed = 90 RPM.
2. Inertia force acts
a) perpendicular to the accelerating force
b) along the direction of accelerating force
c) opposite to the direction of accelerating force
d) none of the mentioned
Explanation: Inertia force acts opposite to the direction of accelerating force
3. The frequency of oscillation at moon compared to earth will be
a) 6 times more
b) 6 times less
c) 2.44 times more
d) 2.44 times less
Explanation: Frequency = 1/2π√g/l
since on moon gravitational force g becomes 1/6g
therefore, frequency = 2.44 times less.
4. Polar moment of inertia(IP) of a circular disc is to be determined by suspending it by a wire and noting the frequency of oscillations(f)
a) IP ∞ f
b) IP ∞ f2
c) IP ∞ 1/f2
d) none of the mentioned
Explanation: IP ∞ 1/f2
5. The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be
a) less
b) more
c) same
d) none of the mentioned
Explanation: The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be more.
The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be less.
6. The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be
a) less
b) more
c) same
d) none of the mentioned
Explanation: The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be more.
The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be less.
7. If the radius of gyration of a compound pendulum about an axis through c.g. is more, then its frequency of oscillation will be
a) less
b) more
c) same
d) none of the mentioned
Explanation: The frequency of oscillation of a bigger diameter cylinder compared to a small cylinder inside a cylinder concave surface will be more.
The frequency of oscillation of a cylinder inside a cylinder inside a cylindrical concave surface of bigger radius compared to a small radius will be less.
8. The Bifilar suspension method is used to determine
a) natural frequency of vibration
b) position of balancing weights
c) moment of inertia
d) none of the mentioned
Explanation: moment of inertia
9. The natural frequency of a spring-mass system on earth is ωn. The natural frequency of this system on the moon (gmoon =gearth/6) is
a) ωn
b) 0.408ωn
c) 0.204ωn
d) 0.167ωn
Explanation: We know natural frequency of a spring mass system is,
ωn = √k/m ………………….(i)
This equation (i) does not depend on the g and weight (W = mg)
So, the natural frequency of a spring mass system is unchanged on the moon.
Hence, it will remain ωn , i.e. ωmoon =ωn.
10. An automotive engine weighing 240 kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 16MN/m while the stiffness of each rear spring is 32MN/m. The engine speed (in rpm), at which resonance is likely to occur, is
a) 6040
b) 3020
c) 1424
d) 955
Explanation: Given k1 = k2 = 16MN/m, k3 = k4 = 32MN/m, m = 240 kg
Here, k1 & k2 are the front two springs or k3 and k4 are the rear two springs.
These 4 springs are parallel, So equivalent stiffness
keq = k1 + k2 + k3 + k4 = 16 + 16 + 32 + 32 = 96MN/m2
We know at resonance
ω = ωn = √k/m
2πN/60 = √keq/m (N =Engine speed in rpm)
N = 60/2π√keq/m
= 60/2π√96 x 106/240
= 6040 rpm.